One of my friend's students attempted to find the volume of a paraboloid using spherical coordinates. I'm trying to see if there's actually a way to finish it. I have been able to find the triple integral which describes the volume of the paraboloid in spherical coordinates and the part that I need help integrating is:
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx.$$
I know that it is equal to $4$ from Wolfram Alpha plus computing the same integral in cylindrical coordinates. Is there a way to finish this from here?
Someone asked for the original problem so here it is. Compute the volume of the region bounded by $z = 1 + x^2 + y^2$ and $z \leq 5$. This is the same as computing the region bounded by $z = 4 – x^2 – y^2$ and the $xy$-plane. Then from there I get the triple integral:
$$\int_{0}^{2\pi}\int_0^{\pi/2}\int_0^{\frac{-\cos{\phi} + \sqrt{\cos^2{(\phi)} + 16\sin^2{(\phi)}}}{2\sin^2{(\phi)}}} \rho^2\sin{\phi}d\rho d\phi d\theta.$$
Of course the integral up top that I care about I just substituted the $\phi$ for an $x$ when I originally stated it by itself.
Best Answer
Let $t=\cos x$, we have \begin{align} I&=\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx \\ &= \frac{1}3\int_0^1 dt \left(\frac{-t+\sqrt{t^2+16(1-t^2)}}{2(1-t^2)}\right)^3 \end{align}
Let $z$ be the larger root of the equation $(1-t^2)z^2+tz-4=0$, we have $z=2$ at $t=0$ and $z=4$ at $t=1$. Let $y=tz$, we have $z^2=y^2-y+4\Rightarrow zdz=(y-1/2)dy$, $y=0$ at $t=0$ and $y=4$ at $t=1$, then
\begin{align} I&=\frac{1}3\int_0^1 dt z(t)^3 \\ &=\frac{1}3\left(z(t)^3t|_0^1-3\int_2^4t(z)z^2dz\right) \\ &=\frac{64}{3}-\int_0^4y\left(y-\frac{1}{2}\right)dy \\ &=\frac{64}{3}-\frac{64}{3}+\frac{16}{4}=4 \end{align}
Also a simple solution to your original problem:
area $A(z)=\pi r(z)^2=\pi(x^2+y^2)=\pi(z-1)$, \begin{align} V&=\int A(z) dz \\ &=\int_1^5\pi(z-1)dz \\ &=8\pi \end{align}